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Question-52208

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Question Number 52208 by SUJIT420 last updated on 04/Jan/19
Commented by maxmathsup by imad last updated on 04/Jan/19
its not c its e .
$${its}\:{not}\:{c}\:{its}\:{e}\:. \\ $$
Commented by maxmathsup by imad last updated on 04/Jan/19
let U(x)=(1+x)^(1/x) −e and V(x)=x we have U(x)=e^((1/x)ln(1+x)) −e ⇒lim_(x→0) U(x)=0 hospital theorem give lim_(x→0) ((U(x))/x) =lim_(x→0) U^′ (x) but U^′ (x)=(((ln(1+x))/x))^′ U(x) =(((x/(1+x))−ln(1+x))/x^2 ) U(x) =((x−(1+x)ln(1+x))/x^2 )U(x) but ln(1+x)∼x ⇒x−(1+x)ln(1+x)∼x−x(1+x)=−x^2 ⇒U^′ (x)∼−U(x) (x→0) we have ln^′ (1+x)=(1/(1+x)) =1−x +o(x^2 ) ⇒ln(1+x)=x−(x^2 /2) +o(x^3 ) ⇒ ((ln(1+x))/x) =1−(x/2) +o(x^2 ) ⇒e^((ln(1+x))/x) =e^(1−(x/2)+o(x^2 )) =e(1−(x/2)+o(x^2 )) ⇒ e^((ln(1+x))/x) −e =−((ex)/2) +o(x^2 )⇒ ((U(x))/x) =−(e/2) +o(x) ⇒lim_(x→0) ((U(x))/x) =−(e/2) and we see that hospital theorem don t work good in this case .
$${let}\:{U}\left({x}\right)=\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} −{e}\:\:{and}\:{V}\left({x}\right)={x}\:{we}\:{have}\: \\ $$$${U}\left({x}\right)={e}^{\frac{\mathrm{1}}{{x}}{ln}\left(\mathrm{1}+{x}\right)} −{e}\:\:\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {U}\left({x}\right)=\mathrm{0}\:\:\:{hospital}\:{theorem}\:{give}\: \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{U}\left({x}\right)}{{x}}\:={lim}_{{x}\rightarrow\mathrm{0}} \:\:{U}^{'} \left({x}\right)\:{but}\:{U}^{'} \left({x}\right)=\left(\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}\right)^{'} {U}\left({x}\right) \\ $$$$=\frac{\frac{{x}}{\mathrm{1}+{x}}−{ln}\left(\mathrm{1}+{x}\right)}{{x}^{\mathrm{2}} }\:{U}\left({x}\right)\:=\frac{{x}−\left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}+{x}\right)}{{x}^{\mathrm{2}} }{U}\left({x}\right)\:{but} \\ $$$${ln}\left(\mathrm{1}+{x}\right)\sim{x}\:\Rightarrow{x}−\left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}+{x}\right)\sim{x}−{x}\left(\mathrm{1}+{x}\right)=−{x}^{\mathrm{2}} \:\Rightarrow{U}^{'} \left({x}\right)\sim−{U}\left({x}\right)\:\left({x}\rightarrow\mathrm{0}\right) \\ $$$${we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\mathrm{1}−{x}\:+{o}\left({x}^{\mathrm{2}} \right)\:\Rightarrow{ln}\left(\mathrm{1}+{x}\right)={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{3}} \right)\:\Rightarrow \\ $$$$\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}\:=\mathrm{1}−\frac{{x}}{\mathrm{2}}\:+{o}\left({x}^{\mathrm{2}} \right)\:\Rightarrow{e}^{\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}} \:={e}^{\mathrm{1}−\frac{{x}}{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right)} \:={e}\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right)\right)\:\Rightarrow \\ $$$${e}^{\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}} −{e}\:\:=−\frac{{ex}}{\mathrm{2}}\:+{o}\left({x}^{\mathrm{2}} \right)\Rightarrow\:\frac{{U}\left({x}\right)}{{x}}\:=−\frac{{e}}{\mathrm{2}}\:+{o}\left({x}\right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{U}\left({x}\right)}{{x}}\:=−\frac{{e}}{\mathrm{2}} \\ $$$$\:{and}\:{we}\:{see}\:{that}\:\:{hospital}\:{theorem}\:{don}\:{t}\:{work}\:{good}\:{in}\:{this}\:{case}\:. \\ $$
Commented by afachri last updated on 05/Jan/19
is it possible to transform (1 + x)^(1/x) − e into Taylor serie Sir ??? because i got stuck to find the derrivative of (1 + x)^(1/x)
$$\mathrm{is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{transform}\:\left(\mathrm{1}\:+\:{x}\right)^{\frac{\mathrm{1}}{{x}}} −\:{e}\:\mathrm{into}\:\mathrm{Taylor} \\ $$$$\mathrm{serie}\:\mathrm{Sir}\:???\:\mathrm{because}\:\mathrm{i}\:\mathrm{got}\:\mathrm{stuck}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{derrivative}\:\mathrm{of}\:\left(\mathrm{1}\:+\:{x}\right)^{\frac{\mathrm{1}}{{x}}} \\ $$
Answered by Smail last updated on 05/Jan/19
lim_(x→0) (((1+x)^(1/x) −e)/x)=lim_(x→0) ((e^((ln(1+x))/x) −e)/x) ln(1+x)≈x−(x^2 /2) near 0 so lim_(x→0) (((1+x)^(1/x) −e)/x)=lim_(x→0) ((e^((x−(x^2 /2))/x) −e)/x) =lim_(x→0) ((e^(1−(x/2)) −e)/x)=lim_(x→0) ((e(e^(−(x/2)) −1))/x) e^t ≈1+t near 0 lim_(x→0) (((1+x)^(1/x) −e)/x)=lim_(x→0) ((e(1−(x/2)−1))/x)=lim_(x→0) ((−xe)/(2x)) =((−e)/2)
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} −{e}}{{x}}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{e}^{\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}} −{e}}{{x}} \\ $$$${ln}\left(\mathrm{1}+{x}\right)\approx{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\:{near}\:\mathrm{0} \\ $$$${so}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} −{e}}{{x}}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{e}^{\frac{{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}{{x}}} −{e}}{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{e}^{\mathrm{1}−\frac{{x}}{\mathrm{2}}} −{e}}{{x}}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{e}\left({e}^{−\frac{{x}}{\mathrm{2}}} −\mathrm{1}\right)}{{x}} \\ $$$${e}^{{t}} \approx\mathrm{1}+{t}\:{near}\:\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} −{e}}{{x}}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{e}\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}−\mathrm{1}\right)}{{x}}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{−{xe}}{\mathrm{2}{x}} \\ $$$$=\frac{−{e}}{\mathrm{2}} \\ $$

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